I think I understand how the DFT can be seen as a bank of bandpass filters. I am trying to understand the IDFT within the same framework. I understand that the IDFT can be seen as summing linear combinations of the complex exponential basis vectors with the coordinates of a signal on that basis but I'd like to understand the filter bank interpretation specifically.

I read that the frequency responses of all DFT channels should sum to N but I don't know how to interpret that fact into a proof that perfect reconstruction is possible via the IDFT formula.

I also read that what matters is the Constant-Overlap-Add value of the window coefficients given a hop size between consecutive transforms (streaming application). Again, I don't know how to use this to derive the IDFT formula. For what its worth, it seems that the DFT would only satisfy the Constant-Overlap-Add condition with a hop size of N.

Can someone explain how the IDFT works as a synthesis filter bank? What conditions are satisfied by the DFT that allow the IDFT formula to reconstruct the signal?


2 Answers 2


"Filterbank" implies "filters", which implies "filtering", or convolution. There is and isn't such equivalence for DFT.


STFT with maximum window width, maximum hop size, and periodic padding, equals DFT.

Bandpasses are compact in frequency by design, either approximately or exactly. The wider the window in time, the narrower it is in frequency. In continuous time, STFT approaches FT in the limit as window width approaches infinity, yielding pure complex sines in time as convolution kernels, and impulses in frequency. window=np.ones(N) is the discrete equivalent of "infinite window". Visually,

For iDFT, we get a nice interpretation via the one-integral inverse: any given timestep of x is recovered by summing rows of STFT at that timestep (with hop_size=1). This works out because STFT is a decomposition, meaning the sum of its parts equals the total.$^{1}$

The standard inversion of STFT involves undoing the convolution: at each timeshift, undo the multiplication by complex sinusoid, and by the window$^{2}$. iDFT then falls out naturally: the first step is undoing $e^{-j\theta}$ with $e^{j\theta} = 1/e^{-j\theta}$.

1: To be strict, the one-integral inverse doesn't exactly work out for general STFT due to imperfect analyticity, but if bandpasses are impulses, they can't "leak over", so this is the one and only exception. An alternative proof is, DFT_matrix[i, :] = DFT_matrix[:, i]. 2: this is actually done at the end, but with the same motivation


Above reasoning is actually circular. The one-integral inverse is proven by showing that the filterbank sums to unity in frequency domain, or sum(DFT(filters)) == 1 for all bins. hop_size=N is also proven by showing equivalence with DFT.

Put simply, DFT is not a convolution, nor does its computation involve convolutions. The very notion of "bandpass" is defined in frequency domain, so "why is DFT made of bandpass filters" is sort of like "why are molecules made of atoms".

In detail, each coefficient of DFT is an inner product that collapses time for each frequency. Each coefficient of one convolution is an inner product that collapses time for each time shift, and the total output is all such time shifts. With hop_size=N, the two are equivalent: each one convolution is only a single coefficient, and the number of convolutions = the number of DFT frequencies. Note, this last equality, n_fft=N, was implicitly assumed throughout the answer: it's the smallest n_fft that satisfies NOLA with hop_size=N.

While DFT != convolution, DFT certainly enables convolution. In trying to visualize the convolution theorem (why "mult in freq" $\Leftrightarrow$ "conv in time"), I found the linearity and time-shift properties to be at its core, a nice view on why convolutions are "the" LTI operators. Along the way I also thought of DFT as convolutions, but I both forgot what it was and it was in a very indirect sense.


Re: invertibility If the goal is to understand why DFT is invertible, there's no need to invoke filtering at all. Some explanations: (1) the DFT is a full rank matrix, $=$ invertible -- (2) the DFT is equivalent to solving a system of $N$ equations.

Further reading: I've not actually read these but they seem relevant: DSP related, Wireless Pi.

Note, this answer looks at STFT as bandpass convolutions / time-frequency transform, rather than "windowed Fourier transform". It is the "truer" interpretation, and more relevant here. If unfamiliar, the CWT visual at top of this answer may help.

Code demo

Standard STFT computes for real x so I show it for rDFT, but full STFT will work for any x. Note, scipy.stft lacks periodic padding.

import numpy as np
from ssqueezepy import stft

N = 64
x = np.random.randn(N)
w = np.ones(N)

ckw = dict(window=w, n_fft=N, padtype='wrap')
Sx = stft(x, **ckw, hop_len=1)
rdft = np.fft.rfft(x)

# validate as DFT (this is same as hop_size==N)
assert np.allclose(rdft, Sx[:, 0])
# validate as one-integral inverse; non-(Nyquist & DC) need adjusting per `.real`
Sx[1:-1] *= 2
x_inv = Sx.sum(axis=0).real / N  # iDFT does / N
assert np.allclose(x_inv, x)

# validate explicitly for hop_size==N
Sx = stft(x, **ckw, hop_len=N).squeeze()
assert np.allclose(rdft, Sx)

I believe that you can split the problem into two parts:

  1. For what window shapes is a «no-operation» overlap-add operation lossless?
  2. Is a FFT-iFFT pair lossless

2 should be evident (ignoring numeric precision) as the DFT and iDFT matrixes are inverses.

1 Amplitude complimentary windows applied twice are a sufficient requirement for lossless overlap add. Some fiddling with window length and the very last sample may be required for even/odd lengths if using commodity window definitions, see this page.

Or more generally JOS have a great overview on this topic: Julius Orion Smith page.

  • $\begingroup$ Added a link to a general JOSIII page... you might want to edit and make it more specific if you have one in mind. $\endgroup$
    – Peter K.
    Jul 16, 2022 at 15:09
  • $\begingroup$ What is the alternative to a no-op overlap add? $\endgroup$
    – B.Ogles
    Jul 16, 2022 at 15:37
  • $\begingroup$ By «no op» I was referring to the unusual requirement that no actual processing is to be carried out between the analysis and synthesis block processing. Usually, you would do something in-between, but as a tool to explore windowing operations, it is fine to do nothing. $\endgroup$
    – Knut Inge
    Jul 16, 2022 at 15:53
  • $\begingroup$ OP appears to ask about the interpretation of iDFT as a convolution filterbank. This post provides nothing to answer that, merely mentioning inversion as a matrix operation, and referencing general filterbank concepts. I don't know why it was accepted. @B.Ogles unless you meant to ask something else, I suggest keeping the question open for other answers. $\endgroup$ Jul 16, 2022 at 18:06
  • 1
    $\begingroup$ @OverLordGoldDragon thanks for the suggestion. I accepted it sort of naively, figuring that maybe the inverse relationship between the DFT/IDFT matrices was all there was to it. I found it a helpful reminder. But you're right that I would appreciate an answer focused on the filter bank interpretation. I have re-opened the question. $\endgroup$
    – B.Ogles
    Jul 16, 2022 at 23:42

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